Determinants — Class 12 Maths Notes

Topics covered in Determinants

1. 1.Introduction to Determinants

   Imagine you're planning a party and have a fixed budget for snacks and decorations. You can choose between two types of snacks and two types of decorations, each with different prices. Determinants can help you decide how many of each to buy without going over your budget. Essentially, determinants are mathematical tools that can be used to solve linear equations, find areas of shapes, and even in more complex applications like understanding how different variables interact in economics, physics, or engineering.
   

   What are Determinants?

   In mathematics, particularly in linear algebra, the determinant is a value that can be computed from the elements of a square matrix. The determinant helps determine the system's behavior, such as whether a system of linear equations has a unique solution, and it is also used in calculating the inverse of a matrix and the area of geometric shapes.

   How Do We Calculate a Determinant?

   For a simple 2x2 matrix, the determinant can be calculated using a straightforward formula. Consider a matrix:

   [][ac​bd​]

   The determinant (let's denote it as ΔΔ) of this matrix is calculated as:

   Δ=−Δ=ad−bc

   This formula subtracts the product of its diagonals. It's like figuring out the balance between two products in a way that affects the whole system.

   Example in Daily Life

   Let's say you have two expenses: snacks (₹a per unit) and decorations (₹c per unit). You spend on ₹b units of snacks and ₹d units of decorations. The determinant can help you understand the total impact of your spending choices on your budget in different scenarios.

   How is This Useful in Real Life and Careers?

   • Engineering: Engineers use determinants to solve systems of linear equations that appear in electrical circuits and mechanical systems.
   • Economics: Economists use determinants when dealing with models that have multiple variables affecting each other, like supply and demand, investment returns.
   • Computer Science: In graphics programming, determinants help in transformations and managing 3D models.

   Activity

   Try calculating the determinant of the following matrix: [3512][31​52​]

   This will give you a feel for how these calculations are performed and can be interpreted.

   Step-by-Step Derivation

   1. Identify the elements of the matrix: =3,=5,=1,=2a=3,b=5,c=1,d=2.
   2. Apply the determinant formula: Δ=−Δ=ad−bc.
   3. Calculate: Δ=(3∗2)−(5∗1)=6−5=1Δ=(3∗2)−(5∗1)=6−5=1.

   The determinant is 1, which in practical scenarios, like solving equations, indicates certain properties about the solutions.

   Conclusion

   Determinants are not just numbers; they're powerful tools that help solve real-world problems where balancing different factors is crucial. By understanding and applying determinants, you're equipping yourself with skills valuable in numerous fields.

2. 2.Determinants of Different Order Matrices

   Determinants provide a unique value calculated from a square matrix and are fundamental in linear algebra for understanding properties like matrix invertibility, solving systems of linear equations, and applications in geometry and physics.

   1. Determinant of a Matrix of Order One (1x1)

   Definition: The determinant of a 1x1 matrix is simply the value of the element itself since there's only one element.

   Matrix Form: =[]A=[a]

   Determinant Calculation: det=det(A)=a

   Examples:

   1. For =[4]A=[4], the determinant is det=4det(A)=4.
   2. For =[−7]A=[−7], the determinant is det=−7det(A)=−7.
   3. For =[0]A=[0], the determinant is det=0det(A)=0.

   These examples demonstrate how straightforward calculating the determinant of a 1x1 matrix is—it’s simply the element itself.

   2. Determinant of a Matrix of Order Two (2x2)

   Definition: The determinant of a 2x2 matrix is calculated using the formula: det=−det(A)=ad−bc where =[]A=[ac​bd​].

   Examples:

   1. =[1324]1. A=[12​34​] det=(1)(4)−(3)(2)=4−6=−2det(A)=(1)(4)−(3)(2)=4−6=−2
   2. =[0110]
      2. A=[01​10​] det=(0)(0)−(1)(1)=−1det(A)=(0)(0)−(1)(1)=−1
   3. =[5231]
      3. A=[53​21​] det=(5)(1)−(2)(3)=5−6=−1det(A)=(5)(1)−(2)(3)=5−6=−1

   These examples illustrate the formula −ad−bc, which effectively balances the products of the matrix's diagonals.

   3. Determinant of a Matrix of Order Three (3x3)

   Definition: The determinant of a 3x3 matrix can be computed using the rule of Sarrus or cofactor expansion, but the rule of Sarrus is a straightforward method involving the sums and differences of the products of diagonals.

   Matrix Form: =[ℎ]A=⎣⎡​adg​beh​cfi​⎦⎤​

   Determinant Calculation (Rule of Sarrus): det=++ℎ−−−ℎdet(A)=aei+bfg+cdh−ceg−bdi−afh

   Examples:

   1. =[123014560]1. A=⎣⎡​105​216​340​⎦⎤​ det=1∗1∗0+2∗4∗5+3∗0∗6−3∗1∗5−2∗0∗0−1∗4∗6=0+40+0−15−0−24=1
      det(A)=1∗1∗0+2∗4∗5+3∗0∗6−3∗1∗5−2∗0∗0−1∗4∗6=0+40+0−15−0−24=1
   2. =[231456789]
      
      2. A=⎣⎡​247​358​169​⎦⎤​ det=2∗5∗9+3∗6∗7+1∗4∗8−1∗5∗7−3∗4∗9−2∗6∗8=90+126+32−35−108−96=9det(A)=2∗5∗9+3∗6∗7+1∗4∗8−1∗5∗7−3∗4∗9−2∗6∗8=90+126+32−35−108−96=9
   3. =[012345111]
      
      3. A=⎣⎡​031​141​251​⎦⎤​ det=0∗4∗1+1∗5∗1+2∗3∗1−2∗4∗1−1∗3∗1−0∗5∗1=0+5+6−8−3−0=0det(A)=0∗4∗1+1∗5∗1+2∗3∗1−2∗4∗1−1∗3∗1−0∗5∗1=0+5+6−8−3−0=0

3. 3.Evaluate the Determinants

   Problem 1:

   Question: Evaluate the determinant: ∣24−5−1∣∣∣​2−5​4−1​∣∣​

   Solution: To find the determinant of a 2x2 matrix, you multiply the top left element by the bottom right element and subtract the product of the top right and bottom left elements. Determinant=(2×−1)−(4×−5)Determinant=(2×−1)−(4×−5) Determinant=−2+20Determinant=−2+20 Determinant=18Determinant=18

   Problem 2:

   (i) Question: Evaluate the determinant: ∣cos⁡−sin⁡sin⁡cos⁡∣∣∣​cosθsinθ​−sinθcosθ​∣∣​

   Solution: The determinant of this 2x2 matrix is found by multiplying the diagonal elements and then subtracting the product of the off-diagonals. Determinant=(cos⁡×cos⁡)−(−sin⁡×sin⁡)Determinant=(cosθ×cosθ)−(−sinθ×sinθ) Using the Pythagorean identity cos⁡2+sin⁡2=1cos2θ+sin2θ=1, Determinant=cos⁡2+sin⁡2Determinant=cos2θ+sin2θ Determinant=1Determinant=1

   (ii) Question: Evaluate the determinant: ∣2−+1−1+1+1∣∣∣​x2−x+1x+1​x−1x+1​∣∣​

   Solution: Again, for a 2x2 matrix: Determinant=(2−+1)(+1)−(−1)(+1)Determinant=(x2−x+1)(x+1)−(x−1)(x+1) Expanding this, we get: Determinant=(3+2−2−++1)−(2+−−1)Determinant=(x3+x2−x2−x+x+1)−(x2+x−x−1) Simplifying by combining like terms: Determinant=3+1−2+1Determinant=x3+1−x2+1 Determinant=3−2+2Determinant=x3−x2+2

   Problem 3:

   Question: If =[1242]A=[14​22​], then show that 2∣∣=4∣∣2∣A∣=4∣A∣.

   Solution: First, let's find the determinant of A: ∣∣=(1×2)−(2×4)∣A∣=(1×2)−(2×4) ∣∣=2−8∣A∣=2−8 ∣∣=−6∣A∣=−6

   Now let's check the property 2∣∣=4∣∣2∣A∣=4∣A∣. Since ∣∣=−6∣A∣=−6, 2∣∣=2×−6=−122∣A∣=2×−6=−12 4∣∣=4×−6=−244∣A∣=4×−6=−24 The property does not hold as stated, it seems there's a mistake in the statement. The correct property should be that if you multiply a 2x2 matrix A by a scalar k, then the determinant of the new matrix is 2k2 times the determinant of A.

   Problem 4:

   Question: If =[101012004]A=⎣⎡​100​010​124​⎦⎤​, then show that 3∣∣=27∣∣3∣A∣=27∣A∣.

   Solution: For a 3x3 matrix A with a scalar k, the determinant of kA is 3k3 times the determinant of A. Let's find the determinant of A: ∣∣=1×∣1204∣∣A∣=1×∣∣​10​24​∣∣​ ∣∣=1×(1×4−2×0)∣A∣=1×(1×4−2×0) ∣∣=4∣A∣=4

   Now, let's verify the property 3∣∣=27∣∣3∣A∣=27∣A∣: 3∣∣=33×∣∣3∣A∣=33×∣A∣ 3∣∣=27×43∣A∣=27×4 3∣∣=1083∣A∣=108 This confirms that 3∣∣=27∣∣3∣A∣=27∣A∣ since 27∣∣27∣A∣ is also 27×4=10827×4=108, which is the property of scalar multiplication on determinants for 3x3 matrices.

4. 4.Find the Determinant

   Problem 2:

   Question: If =[11−221−354−9]A=⎣⎡​125​114​−2−3−9​⎦⎤​, find ∣∣∣A∣.

   Solution: To find the determinant of a 3x3 matrix A, we can use the rule of Sarrus or the expansion by minors. I will use the expansion by minors here.

   ∣∣=1⋅∣1−34−9∣−1⋅∣2−35−9∣+(−2)⋅∣2154∣∣A∣=1⋅∣∣​14​−3−9​∣∣​−1⋅∣∣​25​−3−9​∣∣​+(−2)⋅∣∣​25​14​∣∣​ ∣∣=1⋅((1⋅−9)−(4⋅−3))−1⋅((2⋅−9)−(5⋅−3))+(−2)⋅((2⋅4)−(5⋅1))∣A∣=1⋅((1⋅−9)−(4⋅−3))−1⋅((2⋅−9)−(5⋅−3))+(−2)⋅((2⋅4)−(5⋅1)) ∣∣=1⋅(−9+12)−1⋅(−18+15)−2⋅(8−5)∣A∣=1⋅(−9+12)−1⋅(−18+15)−2⋅(8−5) ∣∣=1⋅3+1⋅(−3)−2⋅3∣A∣=1⋅3+1⋅(−3)−2⋅3 ∣∣=3−3−6∣A∣=3−3−6 ∣∣=−6∣A∣=−6

   Problem 2:

   (i) Question: Find the values of x, if ∣2451∣=∣246∣∣∣​25​41​∣∣​=∣∣​2x6​4x​∣∣​

   Solution: First, we calculate the determinant on the left side: Left Det=(2⋅1)−(4⋅5)Left Det=(2⋅1)−(4⋅5) Left Det=2−20Left Det=2−20 Left Det=−18Left Det=−18

   Now, we equate it to the determinant on the right side and solve for x: Right Det=(2⋅)−(4⋅6)Right Det=(2x⋅x)−(4⋅6) 22−24=−182x2−24=−18 22=62x2=6 2=3x2=3 =±3x=±3​

   (ii) Question: Find the values of x, if ∣2345∣=∣325∣∣∣​24​35​∣∣​=∣∣​x2x​35​∣∣​

   Solution: First, we calculate the determinant on the left side: Left Det=(2⋅5)−(3⋅4)Left Det=(2⋅5)−(3⋅4) Left Det=10−12Left Det=10−12 Left Det=−2Left Det=−2

   Now, equate it to the determinant on the right side and solve for x: Right Det=(⋅5)−(3⋅2)Right Det=(x⋅5)−(3⋅2x) 5−6=−25x−6x=−2 −=−2−x=−2 =2x=2

5. 5.Area of a Triangle

   The area of a triangle with vertices at points (1,1)(x1​,y1​), (2,2)(x2​,y2​), and (3,3)(x3​,y3​) can be found using determinants by the following formula:

   Area=12∣∣111221331∣∣Area=21​∣∣​∣∣​x1​x2​x3​​y1​y2​y3​​111​∣∣​∣∣​

   This is known as the determinant formula for the area of a triangle. To apply this, you simply substitute the x and y coordinates of the triangle's vertices into the determinant and calculate its absolute value (to ensure the area is positive) and then multiply by 1221​ because the determinant actually gives you twice the area of the triangle.

   Here’s a step-by-step way to do it:

   1. Place the coordinates of the three vertices into the first and second columns of a 3x3 matrix.
   2. Fill the third column with the number 1.
   3. Calculate the determinant of this matrix.
   4. Take the absolute value of the determinant and then divide by 2.

   Let's go through an example. Suppose you have a triangle with vertices at (1,2)(1,2), (3,4)(3,4), and (5,0)(5,0). The matrix would be:

   ∣121341501∣∣∣​135​240​111​∣∣​

   Now we calculate the determinant:

   =∣(1⋅4⋅1+2⋅1⋅5+1⋅3⋅0)−(1⋅3⋅1+1⋅4⋅5+2⋅1⋅0)∣=∣(1⋅4⋅1+2⋅1⋅5+1⋅3⋅0)−(1⋅3⋅1+1⋅4⋅5+2⋅1⋅0)∣ =∣(4+10+0)−(3+20+0)∣=∣(4+10+0)−(3+20+0)∣ =∣14−23∣=∣14−23∣ =∣−9∣=∣−9∣

   So the area of the triangle is 12×9=4.521​×9=4.5 square units.

6. 6.Find Area of the Triangle with vertices at the point given in each of the following :

   1. For vertices (1, 0), (6, 0), (4, 3):

   The determinant of the matrix formed by the vertices is:

   ∣101601431∣∣∣​164​003​111​∣∣​

   We calculate it as follows:

   =1(0⋅1−3⋅1)−0(6⋅1−3⋅4)+1(6⋅3−0)=1(0⋅1−3⋅1)−0(6⋅1−3⋅4)+1(6⋅3−0) =1(0−3)−0+18=1(0−3)−0+18 =−3+18=−3+18 =15=15

   So, the area is 12×∣15∣=7.521​×∣15∣=7.5 square units.

   2. For vertices (2, 7), (1, 1), (10, 8):

   The determinant of the matrix formed by the vertices is:

   ∣2711111081∣∣∣​2110​718​111​∣∣​

   Calculating it, we get:

   =2(1⋅1−8⋅1)−7(1⋅1−10⋅1)+1(1⋅8−7⋅1)=2(1⋅1−8⋅1)−7(1⋅1−10⋅1)+1(1⋅8−7⋅1) =2(1−8)−7(1−10)+(8−7)=2(1−8)−7(1−10)+(8−7) =2(−7)−7(−9)+1=2(−7)−7(−9)+1 =−14+63+1=−14+63+1 =50=50

   Thus, the area is 12×∣50∣=2521​×∣50∣=25 square units.

   3. For vertices (−2, −3), (3, 2), (−1, −8):

   The determinant of the matrix formed by the vertices is:

   ∣−2−31321−1−81∣∣∣​−23−1​−32−8​111​∣∣​

   We calculate it as:

   =(−2)(2⋅1−(−8)⋅1)−(−3)(3⋅1−(−1)⋅1)+1(3⋅(−8)−2⋅(−1))=(−2)(2⋅1−(−8)⋅1)−(−3)(3⋅1−(−1)⋅1)+1(3⋅(−8)−2⋅(−1)) =(−2)(2+8)+3(3+1)−(24−2)=(−2)(2+8)+3(3+1)−(24−2) =(−2)(10)+3(4)−22=(−2)(10)+3(4)−22 =−20+12−22=−20+12−22 =−30=−30

   So, the area is 12×∣−30∣=1521​×∣−30∣=15 square units.

7. 7.Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear

   To show that points (,+)A(a,b+c), (,+)B(b,c+a), and (,+)C(c,a+b) are collinear, we can use the concept of determinants. If the area of the triangle formed by three points is zero, then these points are collinear.

   We form a matrix using the coordinates of the points A, B, and C as rows and calculate its determinant. If the determinant is zero, it implies that the area of the triangle is zero, and hence the points are collinear.

   Let's construct the determinant:

   ∣+1+1+1∣∣∣​abc​b+cc+aa+b​111​∣∣​

   We expand the determinant:

   =((+)⋅1−(+)⋅1)−(+)(⋅1−⋅1)+1(⋅(+)−⋅(+))=a((c+a)⋅1−(a+b)⋅1)−(b+c)((b)⋅1−(c)⋅1)+1((b)⋅(a+b)−(c)⋅(c+a)) =(+−−)−(+)(−)+(+2−2−)=a(c+a−a−b)−(b+c)(b−c)+(ab+b2−c2−ac) =(−)−(2−2+−)+(+2−2−)=a(c−b)−(b2−c2+bc−bc)+(ab+b2−c2−ac) =−−2+2++2−2+=ac−ab−b2+c2+ab+b2−c2+ac =2−2=2ac−2ac =0=0

   The determinant is zero, which implies that the points A, B, and C are collinear.

8. 8.Find values of k if area of triangle is 4 sq. units and vertices are..(i) (k, 0), (4, 0), (0, 2) , (ii) (–2, 0), (0, 4), (0, k)

   To find the values of k for which the area of the triangle is 4 square units, we'll again use the determinant method. The area A of a triangle with vertices (1,1)(x1​,y1​), (2,2)(x2​,y2​), and (3,3)(x3​,y3​) is given by:

   =12∣1(2−3)+2(3−1)+3(1−2)∣A=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

   For the given area to be 4 sq. units, the absolute value of the determinant must be 8 because the area is half of the determinant.

   Let's solve each part:

   (i) For vertices (k, 0), (4, 0), (0, 2):

   8=∣(0−2)+4(2−0)+0(0−0)∣8=∣k(0−2)+4(2−0)+0(0−0)∣ 8=∣−2+8∣8=∣−2k+8∣

   This equation will give us the possible values for k. We solve for k:

   −2+8=8−2k+8=8 −2=0−2k=0 =0k=0

   Or

   −2+8=−8−2k+8=−8 −2=−16−2k=−16 =8k=8

   So, for the first case, k can be either 0 or 8.

   (ii) For vertices (–2, 0), (0, 4), (0, k):

   8=∣(−2)(4−)+0(−0)+0(0−4)∣8=∣(−2)(4−k)+0(k−0)+0(0−4)∣ 8=∣−8+2∣8=∣−8+2k∣

   Again, we solve for k:

   −8+2=8−8+2k=8 2=162k=16 =8k=8

   Or

   −8+2=−8−8+2k=−8 2=02k=0 =0k=0

   In the second case, k can also be either 0 or 8.

   Therefore, the values of k that result in an area of 4 square units for the given vertices in both cases are 0 and 8.

9. 9.(i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

   To find the equation of the line joining two points using determinants, we can use the fact that the area of the triangle formed by the line and a third point (,)(x,y) is zero if (,)(x,y) lies on the line. This leads us to the following determinant being zero:

   ∣1112211∣=0∣∣​x1​x2​x​y1​y2​y​111​∣∣​=0

   Let's find the equations:

   (i) For the line joining (1, 2) and (3, 6):

   We plug the coordinates of the points into the determinant to get the line equation:

   ∣1213611∣=0∣∣​13x​26y​111​∣∣​=0

   Expanding the determinant:

   (1⋅6⋅1+2⋅1⋅+1⋅3⋅)−(1⋅3⋅1+1⋅6⋅+2⋅1⋅)=0(1⋅6⋅1+2⋅1⋅x+1⋅3⋅y)−(1⋅3⋅1+1⋅6⋅x+2⋅1⋅y)=0

   6+2+3−3−6−2=06+2x+3y−3−6x−2y=0

   Simplifying:

   2−6+3−2+6−3=02x−6x+3y−2y+6−3=0

   −4++3=0−4x+y+3=0

   Therefore, the equation of the line is =4−3y=4x−3.

   (ii) For the line joining (3, 1) and (9, 3):

   Similarly, we use the determinant:

   ∣3119311∣=0∣∣​39x​13y​111​∣∣​=0

   Expanding the determinant:

   (3⋅3⋅1+1⋅1⋅+1⋅9⋅)−(1⋅9⋅1+1⋅3⋅+3⋅1⋅)=0(3⋅3⋅1+1⋅1⋅x+1⋅9⋅y)−(1⋅9⋅1+1⋅3⋅x+3⋅1⋅y)=0

   9++9−9−3−3=09+x+9y−9−3x−3y=0

   Simplifying:

   −3+9−3=0x−3x+9y−3y=0

   −2+6=0−2x+6y=0

   =13y=31​x

   Therefore, the equation of the line is =13y=31​x.

10. 10.Minors and Cofactors

    Minors and cofactors are concepts from linear algebra, particularly relevant in the context of matrices. They are essential in calculating the determinant of a matrix and in finding the inverse of a matrix.

    Minors: The minor of an element in a matrix is the determinant of the smaller matrix that remains after removing the row and column of that element. It is typically denoted as Mij​, where i and j represent the row and column indices of the element.

    Cofactors: The cofactor of an element in a matrix is the minor of that element, multiplied by (−1)+(−1)i+j, which is known as the cofactor sign factor. The cofactor is denoted as Cij​.

    Let's see some examples of how to calculate minors and cofactors.

    Example 1: Consider the matrix A given by =[123456789]A=⎣⎡​147​258​369​⎦⎤​

    To find the minor of element 22a22​ (which is the element 5 in this case), we remove the second row and the second column: 22=∣1379∣=(1⋅9)−(3⋅7)=9−21=−12M22​=∣∣​17​39​∣∣​=(1⋅9)−(3⋅7)=9−21=−12

    The cofactor of element 22a22​ is: 22=(−1)2+2⋅22=1⋅(−12)=−12C22​=(−1)2+2⋅M22​=1⋅(−12)=−12

    Example 2: Now, let's calculate the minor and cofactor of element 13a13​ (which is the element 3).

    The minor of 13a13​ is: 13=∣4578∣=(4⋅8)−(5⋅7)=32−35=−3M13​=∣∣​47​58​∣∣​=(4⋅8)−(5⋅7)=32−35=−3

    The cofactor of element 13a13​ is: 13=(−1)1+3⋅13=−1⋅(−3)=3C13​=(−1)1+3⋅M13​=−1⋅(−3)=3

    Example 3: Finally, let's find the minor and cofactor for element 31a31​, which is the element 7.

    The minor of 31a31​ is: 31=∣2356∣=(2⋅6)−(3⋅5)=12−15=−3M31​=∣∣​25​36​∣∣​=(2⋅6)−(3⋅5)=12−15=−3

    The cofactor of element 31a31​ is: 31=(−1)3+1⋅31=1⋅(−3)=−3C31​=(−1)3+1⋅M31​=1⋅(−3)=−3

11. 11.Write Minors and Cofactors of the elements of following determinants:

    Question 1:

    (i) For the determinant ∣2−403∣∣∣​20​−43​∣∣​, calculate the minors and cofactors.

    Solution 1:

    The minors for each element are:

    • Minor of 2 (11M11​): 33 (Determinant of the matrix after removing the row and column of the element 2)
    • Minor of -4 (12M12​): 00 (Determinant of the matrix after removing the row and column of the element -4)
    • Minor of 0 (21M21​): −4−4 (Determinant of the matrix after removing the row and column of the element 0)
    • Minor of 3 (22M22​): 22 (Determinant of the matrix after removing the row and column of the element 3)

    The cofactors for each element are:

    • Cofactor of 2 (11C11​): (−1)1+1⋅11=1⋅3=3(−1)1+1⋅M11​=1⋅3=3
    • Cofactor of -4 (12C12​): (−1)1+2⋅12=−1⋅0=0(−1)1+2⋅M12​=−1⋅0=0
    • Cofactor of 0 (21C21​): (−1)2+1⋅21=−1⋅(−4)=4(−1)2+1⋅M21​=−1⋅(−4)=4
    • Cofactor of 3 (22C22​): (−1)2+2⋅22=1⋅2=2(−1)2+2⋅M22​=1⋅2=2

    (2) For the determinant ∣∣∣∣​ab​cd​∣∣​, calculate the minors and cofactors.

    Solution 2:

    The minors for each element are:

    • Minor of a (11M11​): d
    • Minor of c (12M12​): b
    • Minor of b (21M21​): c
    • Minor of d (22M22​): a

    The cofactors for each element are:

    • Cofactor of a (11C11​): (−1)1+1⋅11=1⋅=(−1)1+1⋅M11​=1⋅d=d
    • Cofactor of c (12C12​): (−1)1+2⋅12=−1⋅=−(−1)1+2⋅M12​=−1⋅b=−b
    • Cofactor of b (21C21​): (−1)2+1⋅21=−1⋅=−(−1)2+1⋅M21​=−1⋅c=−c
    • Cofactor of d (22C22​): (−1)2+2⋅22=1⋅=(−1)2+2⋅M22​=1⋅a=a

12. 12.Using Cofactors of elements of second row,..

    Question:

    Using cofactors of elements of the second row, evaluate ΔΔ for the matrix:

    Δ=∣538201123∣Δ=∣∣​521​302​813​∣∣​

    Solution:

    The determinant of a matrix can be found by multiplying each element of any row or column with its corresponding cofactor and adding them up. We will use the second row here.

    First, let's calculate the cofactors for the second row:

    • The cofactor of the element 2 (21C21​) is given by:

    21=(−1)2+1⋅∣3823∣=−((3⋅3)−(8⋅2))=−(9−16)=7C21​=(−1)2+1⋅∣∣​32​83​∣∣​=−((3⋅3)−(8⋅2))=−(9−16)=7

    • The cofactor of the element 0 (22C22​) is given by:

    22=(−1)2+2⋅∣5813∣=(5⋅3)−(8⋅1)=15−8=7C22​=(−1)2+2⋅∣∣​51​83​∣∣​=(5⋅3)−(8⋅1)=15−8=7

    • The cofactor of the element 1 (23C23​) is given by:

    23=(−1)2+3⋅∣5312∣=−((5⋅2)−(3⋅1))=−(10−3)=−7C23​=(−1)2+3⋅∣∣​51​32​∣∣​=−((5⋅2)−(3⋅1))=−(10−3)=−7

    Now we can find ΔΔ using the second row and its cofactors:

    Δ=2⋅21+0⋅22+1⋅23Δ=2⋅C21​+0⋅C22​+1⋅C23​ Δ=2⋅7+0⋅7+1⋅(−7)Δ=2⋅7+0⋅7+1⋅(−7) Δ=14−7Δ=14−7 Δ=7Δ=7

13. 13.Using Cofactors of elements of third column,

    Question:

    Using cofactors of elements of the third column, evaluate ΔΔ for the matrix:

    Δ=∣∣Δ=∣∣​xyz​yzx​yzzxxy​∣∣​

    Solution for Question 4:

    Let's calculate the cofactors for the third column:

    • The cofactor of the element yz (13C13​) is given by:

    13=(−1)1+3⋅∣∣=−((⋅)−(⋅))=−(−2)C13​=(−1)1+3⋅∣∣​yz​zx​∣∣​=−((y⋅x)−(z⋅z))=−(yx−z2)

    • The cofactor of the element zx (23C23​) is given by:

    23=(−1)2+3⋅∣∣=((⋅)−(⋅))=2−C23​=(−1)2+3⋅∣∣​xz​yx​∣∣​=((x⋅x)−(y⋅z))=x2−yz

    • The cofactor of the element xy (33C33​) is given by:

    33=(−1)3+3⋅∣∣=(⋅)−(⋅)=−2C33​=(−1)3+3⋅∣∣​xy​yz​∣∣​=(x⋅z)−(y⋅y)=xz−y2

    Now we can find ΔΔ using the third column and its cofactors:

    Δ=⋅13+⋅23+⋅33Δ=yz⋅C13​+zx⋅C23​+xy⋅C33​ Δ=⋅(−(−2))+⋅(2−)+⋅(−2)Δ=yz⋅(−(yx−z2))+zx⋅(x2−yz)+xy⋅(xz−y2) Δ=−2+2+3−2+2−3Δ=−yz2+y2zx+zx3−zxy2+xyz2−xy3 Δ=3−3Δ=zx3−xy3 (Since −2+2=0−yz2+xyz2=0 and 2−2=0y2zx−zxy2=0)

    We can further simplify the determinant if there are common factors or if the expression can be factored, but based on the elements provided, the determinant simplifies to 3−3zx3−xy3.

14. 14.Adjoint and Inverse of a Matrix and It's Theoram

    Adjoint of a Matrix:

    Definition Expansion: For an ×n×n square matrix A, the adjoint is found by taking the transpose of the cofactor matrix of A.

    Extended Example for Adjoint: Consider a 3x3 matrix A: =[123045106]A=⎣⎡​101​240​356​⎦⎤​

    To find the adjoint, we calculate the cofactor for each element in the matrix. For instance, the cofactor 11C11​ for element 11a11​ is given by: 11=(−1)1+1∣4506∣=(4)(6)−(5)(0)=24C11​=(−1)1+1∣∣​40​56​∣∣​=(4)(6)−(5)(0)=24

    Calculating similarly, we get the cofactor matrix: =[24−5−406−60−54]C=⎣⎡​2400​−56−5​−4−64​⎦⎤​

    Taking the transpose of the cofactor matrix C, we get the adjoint of A: adj=[2400−56−5−4−64]adj(A)=⎣⎡​24−5−4​06−6​0−54​⎦⎤​

    Inverse of a Matrix:

    Definition Expansion: For an ×n×n square matrix A, the inverse of A, if it exists, is a matrix −1A−1 such that ⋅−1=A⋅A−1=In​ and −1⋅=A−1⋅A=In​, where In​ is the identity matrix of size n.

    Extended Example for Inverse: Using the same matrix A and its adjoint adjadj(A) from the previous example, let’s find the inverse. We calculate the determinant of A first: det⁡=1∣4506∣−2∣0516∣+3∣0410∣=(24+10−12)=22det(A)=1∣∣​40​56​∣∣​−2∣∣​01​56​∣∣​+3∣∣​01​40​∣∣​=(24+10−12)=22

    Since the determinant is non-zero, A is invertible. Now, the inverse is given by: −1=1det⁡adj=122[2400−56−5−4−64]A−1=det(A)1​adj(A)=221​⎣⎡​24−5−4​06−6​0−54​⎦⎤​

    Theorems Related to Adjoint and Inverse:

    Theorem 1 Expansion: If A is an invertible matrix, then ⋅adj=adj⋅=det⁡⋅A⋅adj(A)=adj(A)⋅A=det(A)⋅I. This relationship underlines the process of finding an inverse via the adjoint.

    Theorem 1 Example: Using our previous matrix A and adjoint adjadj(A), we can demonstrate this theorem: ⋅adj=[123045106]⋅[2400−56−5−4−64]=[220002200022]A⋅adj(A)=⎣⎡​101​240​356​⎦⎤​⋅⎣⎡​24−5−4​06−6​0−54​⎦⎤​=⎣⎡​2200​0220​0022​⎦⎤​ which is 22⋅322⋅I3​, confirming the theorem.

    Theorem 2 Expansion: If A and B are invertible matrices, then −1=−1−1(AB)−1=B−1A−1. This theorem is essential when dealing with multiple linear transformations.

    Theorem 2 Example: If we have another matrix =[78910]B=[79​810​] with its inverse −1B−1, and we compute AB and then find −1(AB)−1, it will equal −1−1B−1A−1.

    Theorem 3 Expansion: The inverse of the transpose of an invertible matrix A is the transpose of the inverse, i.e., −1=(−1)(AT)−1=(A−1)T. This is often used in statistical analysis and in solving linear equation systems.

    Theorem 3 Example: For any invertible matrix A, if we compute AT and then find −1(AT)−1, it will equal the transpose of −1A−1, which we found above.

    Theorem 4 Expansion: If A is an invertible matrix, then the adjoint of A can be expressed in terms of its determinant and inverse: adj=det⁡⋅−1adj(A)=det(A)⋅A−1.

    Theorem 4 Example: Using our matrix A and its determinant and inverse from above, we can verify that: adj=det⁡⋅−1adj(A)=det(A)⋅A−1

    Applications in Real Life and Careers:

    1. Engineering: When designing systems or structures, engineers often need to solve sets of linear equations to determine force distributions, electrical currents, or material stresses. This process can be greatly facilitated by using matrix inverses.

    2. Computer Science: In graphics programming, transformations like rotations, scaling, and translations are represented by matrices. Adjusting these transformations requires computing the inverse of transformation matrices.

    3. Economics: Economists use matrices to model economic activities across different sectors. The Leontief inverse, which is a matrix inverse, is crucial in input-output analysis for predicting the effect of changes in one industry on others.

    4. Physics: In quantum mechanics, the adjoint of an operator matrix relates to the concept of hermitian operators, which are fundamental in the formulation of physical theories.

    In each of these fields, a strong grasp of matrices and their properties like adjoint and inverse is crucial.

15. 15.Some Exercises Questions

    Question 1: Find the adjoint of the matrix:
    

    =[1234]A=[13​24​]

    Solution: To find the adjoint of a 2x2 matrix, we swap the elements on the main diagonal and change the signs of the off-diagonal elements. adj=[4−3−21]adj(A)=[4−2​−31​]

    Question 2: Find the adjoint of the matrix: =[1−12235−201]B=⎣⎡​12−2​−130​251​⎦⎤​

    Solution: The adjoint of a 3x3 matrix is obtained by finding the cofactor of each element and then transposing the cofactor matrix. adj=[3−126152−11−15]adj(B)=⎣⎡​31−11​−125−1​625​⎦⎤​

    Question 3: Verify (adj)=(adj)=∣∣A(adjA)=(adjA)A=∣A∣I for the matrix: =[23−4−6]C=[2−4​3−6​]

    Solution: First, we find the determinant of C, ∣∣∣C∣, and then the adjoint, adjadj(C). Finally, we multiply C with adjadj(C) and verify the equality.

    Question4.: Verify
    

    (adj)=(adj)=∣∣A(adjA)=(adjA)A=∣A∣I for the matrix: =[1−1230−2103]D=⎣⎡​131​−100​2−23​⎦⎤​

    Solution: Similarly, we find ∣∣∣D∣, adjadj(D), and then multiply D with adjadj(D) to verify the equality.

    Question 5: Find the inverse of the matrix (if it exists):
    

    =[2−243]E=[24​−23​]

    Solution: The inverse is given by −1=1det⁡adjE−1=det(E)1​adj(E). We need to calculate the determinant of E and then multiply the reciprocal of that determinant with the adjoint of E.

    Question 6: Find the inverse of the matrix (if it exists):
    

    =[−15−32]F=[−1−3​52​]

    Solution: We use the formula −1=1det⁡adjF−1=det(F)1​adj(F), similar to Exercise 5, by calculating the determinant and adjoint of F.

    Question 7: Find the inverse of the matrix (if it exists):
    

    =[123024005]G=⎣⎡​100​220​345​⎦⎤​

    Solution: Since G is an upper triangular matrix, its inverse is easier to calculate. The determinant is the product of the diagonal elements. The inverse −1G−1 can be found by taking the reciprocal of the diagonal elements and performing necessary operations on the off-diagonal elements.

    Here are the detailed solutions for Question 3 to 7:
    

    Question 3:

    Determinant of C: ∣∣=0∣C∣=0

    Adjoint of C: adj=[−6−342]adj(C)=[−64​−32​]

    Verification: Since the determinant is zero, the product ⋅adjC⋅adj(C) equals a zero matrix, which verifies the identity (adj)=(adj)=∣∣A(adjA)=(adjA)A=∣A∣I.

    Question 4:

    Determinant of D: ∣∣=11∣D∣=11

    Adjoint of D: adj=[032−11180−13]adj(D)=⎣⎡​0−110​31−1​283​⎦⎤​

    Verification: The product ⋅adjD⋅adj(D) equals 1111 times the identity matrix, verifying the identity (adj)=(adj)=∣∣A(adjA)=(adjA)A=∣A∣I.

    Question 5:

    Inverse of E: −1=[31417−2717]E−1=[143​−72​​71​71​​]

    Question 6:

    Inverse of F: −1=[213−513313−113]F−1=[132​133​​−135​−131​​]

    Question 7:

    Inverse of G: −1=[1−115012−250015]G−1=⎣⎡​100​−121​0​51​−52​51​​⎦⎤​